The lifespans of bears in a particular zoo are normally distributed. The average bear lives $47.6$ years; the standard deviation is $10.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $58.2$ years.
$47.6$ $37$ $58.2$ $26.4$ $68.8$ $15.8$ $79.4$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $47.6$ years. We know the standard deviation is $10.6$ years, so one standard deviation below the mean is $37$ years and one standard deviation above the mean is $58.2$ years. Two standard deviations below the mean is $26.4$ years and two standard deviations above the mean is $68.8$ years. Three standard deviations below the mean is $15.8$ years and three standard deviations above the mean is $79.4$ years. We are interested in the probability of a bear living less than $58.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $37$ years and the other half $({16\%})$ will live longer than $58.2$ years. The probability of a particular bear living less than $58.2$ years is ${68\%} + {16\%}$, or $84\%$.